Free Books
   Free Books    Introduction to Digital Filters  

Unstable Poles--Unit Circle Viewpoint

We saw in §8.4 that an LTI filter is stable if and only if all of its poles are strictly inside the unit circle ($ \vert z\vert=1$) in the complex $ z$ plane. In particular, a pole $ p$ outside the unit circle ($ \vert p\vert>1$) gives rise to an impulse-response component proportional to $ p^n$ which grows exponentially over time $ n$. We also saw in §6.2 that the z transform of a growing exponential does not not converge on the unit circle in the $ z$ plane. However, this was the case for a causal exponential $ u(n)p^n$, where $ u(n)$ is the unit-step function (which switches from 0 to 1 at time 0). If the same exponential is instead anticausal, i.e., of the form $ u(-n)p^n$, then, as we'll see in this section, its z transform does exist on the unit circle, and the pole is in exactly the same place as in the causal case. Therefore,to unambiguously invert a z transform, we must know its region of convergence. The critical question is whether the region of convergence includes the unit circle: If it does, then each pole outside the unit circle corresponds to an anticausal, finite energy, exponential, while each pole inside corresponds to the usual causal decaying exponential.

Geometric Series

The essence of the situation can be illustrated using a simple geometric series. Let $ R$ be any real (or complex) number. Then we have

$\displaystyle \frac{1}{1-R} \eqsp 1 + R + R^2 + R^3 + \cdots \quad < \infty$   when$\displaystyle \quad\vert R\vert<1. $

In other words, the geometric series $ 1 + R + R^2 + R^3 + \cdots$ is guaranteed to be summable when $ \vert R\vert<1$, and in that case, the sum is given by $ 1/(1-R)$. On the other hand, if $ \vert R\vert>1$, we can rewrite $ 1/(1-R)$ as $ -R^{-1}/(1-R^{-1})$ to obtain

$\displaystyle \frac{1}{1-R} \eqsp \frac{-R^{-1}}{1-R^{-1}} \eqsp -R^{-1}\left[1 + R^{-1} + R^{-2} + R^{-3} + \cdots \right] $

which is summable when $ \vert R\vert>1$. Thus, $ 1/(1-R)$ is a valid closed-form sum whether or not $ \vert R\vert$ is less than or greater than 1. When $ \vert R\vert<1$, it is the sum of the causal geometric series in powers of $ R$. When $ \vert R\vert>1$, it is the sum of the causal geometric series in powers of $ R^{-1}$, or, an anticausal geometric series in (negative) powers of $ R$.

One-Pole Transfer Functions

We can apply the same analysis to a one-pole transfer function. Let $ p\in{\bf C}$ denote any real or complex number:

$\displaystyle H(z) \eqsp \frac{1}{1-pz^{-1}} \eqsp 1 + pz^{-1}+ pz^{-2}+ pz^{-3} + \cdots $

The convergence criterion is now $ \vert pz^{-1}\vert<1$, or $ \vert z\vert>\vert p\vert$. For the region of convergence to include the unit circle (our frequency axis), we must have $ \vert p\vert<1$, which is our usual stability criterion for a pole at $ z=p$. The inverse z transform is then the causal decaying sampled exponential

$\displaystyle H(z) \;\longleftrightarrow\; h(n) = u(n)p^n $

Now consider the rewritten case:

\begin{eqnarray*} \frac{1}{1-pz^{-1}} &=& \frac{-p^{-1}z}{1-p^{-1}z} \\ &=& -p^... ...cdots\right]\\ &\leftrightarrow& - u(-n-1)p^n,\quad n\in{\bf Z} \end{eqnarray*}

where the inverse z transform is the inverse bilateral z transform. In this case, the convergence criterion is $ \vert p^{-1}z\vert<1$, or $ \vert z\vert<\vert p\vert$, and this region includes the unit circle when $ \vert p\vert>1$.

In summary, when the region-of-convergence of the z transform is assumed to include the unit circle of the $ z$ plane, poles inside the unit circle correspond to stable, causal, decaying exponentials, while poles outside the unit circle correspond to anticausal exponentials that decay toward time $ -\infty$, and stop before time zero.

Figure 8.8 illustrates the two types of exponentials (causal and anticausal) that correspond to poles (inside and outside the unit circle) when the z transform region of convergence is defined to include the unit circle.

myFourFiguresToWidthpolesout11polesout21polesout12polesout220.52Left column: Causal exponential decay, pole at $ p=0.9$. Right column: Anticausal exponential decay, pole at $ p=1/0.9$. Top: Pole-zero diagram. Bottom: Corresponding impulse response, assuming the region of convergence includes the unit circle in the $ z$ plane.

Next Section:
Poles and Zeros of the Cepstrum
Previous Section:
Time Constant of One Pole